Deriving the Bivariate Normal Density

Why Derive the Density?

For a bivariate normal pair, the correlation affects more than the direction of association: it changes the shape of the joint density and appears in both its exponent and normalizing constant. Deriving the density makes those roles explicit and provides a reliable way to evaluate it without losing a sign or a scale factor.

This tutorial concerns the nondegenerate bivariate normal model. A density exists with respect to area in the plane only when the correlation is strictly between $-1$ and $1$. The derivation also shows why the exponent is a quadratic expression in the standardized coordinates, rather than the sum of two independent squared deviations.

Definition: A nondegenerate bivariate normal pair $(X,Y)$ has means $\mu_X,\mu_Y$, positive standard deviations $\sigma_X,\sigma_Y$, and correlation $\rho$ satisfying $-1<\rho<1$. Its covariance is $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y$. Define the standardized coordinates $u=(x-\mu_X)/\sigma_X$ and $v=(y-\mu_Y)/\sigma_Y$. The joint density is $$ f_{X,Y}(x,y)=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}} \exp\left[-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right]. $$

The parameters have distinct roles. The means locate the center of the density; the standard deviations set the horizontal and vertical scales; and $\rho$ controls how the two standardized coordinates vary together. The denominator $2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}$ is the normalizing factor that makes the density integrate to one over the whole plane.

The formula applies when the pair is bivariate normal, both standard deviations are positive, and $|\rho|<1$. It is not a general density formula for any pair of variables that happen to have those means, variances, and correlation. At $\rho=1$ or $\rho=-1$, the distribution is degenerate, concentrated on a line, and does not have an ordinary two-dimensional density.

Derivation From Independent Standard Normals

Let $Z_1$ and $Z_2$ be independent standard normal variables, each with mean zero and variance one. Write $s=\sqrt{1-\rho^2}$, which is positive because $|\rho|<1$. Construct the pair by $$ X=\mu_X+\sigma_X Z_1,\qquad Y=\mu_Y+\sigma_Y\left(\rho Z_1+sZ_2\right). $$

The standardized coordinates then satisfy $u=Z_1$ and $v=\rho Z_1+sZ_2$. In particular, $Z_2=(v-\rho u)/s$. The transformation from $(Z_1,Z_2)$ to $(X,Y)$ has a triangular Jacobian matrix, whose determinant is $\sigma_X\sigma_Ys$. Its absolute value is the area-scaling factor needed in the change-of-variables formula.

Write the independent density.
Independence gives the joint density of $Z_1,Z_2$ as $(2\pi)^{-1}\exp[-(z_1^2+z_2^2)/2]$.

Express the old coordinates in terms of the new ones.
Use $z_1=u$ and $z_2=(v-\rho u)/s$, where $s=\sqrt{1-\rho^2}$.

Apply the Jacobian adjustment.
Divide the standard-normal joint density by $\sigma_X\sigma_Ys$, the absolute determinant of the transformation from standard-normal coordinates to $(x,y)$.

Expand and simplify the exponent.
Combine the squared terms to obtain the correlated quadratic form and the stated density.

For completeness, substitution into the independent density gives the exponent through the sum $z_1^2+z_2^2$. Expanding that sum shows exactly where the cross-product term comes from:

z_1^2+z_2^2 =u^2+\frac{(v-\rho u)^2}{1-\rho^2} =\frac{u^2-2\rho uv+v^2}{1-\rho^2}.

The final equality follows by multiplying the first expression by $1-\rho^2$: the numerator becomes $u^2(1-\rho^2)+v^2-2\rho uv+\rho^2u^2=u^2-2\rho uv+v^2$. The change-of-variables formula therefore gives

f_{X,Y}(x,y) =\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}} \exp\left[-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right].

Why the normalizing constant has this form: The independent standard normal density contributes $1/(2\pi)$. Transforming to the $(x,y)$ scale divides the density by the absolute Jacobian $\sigma_X\sigma_Y\sqrt{1-\rho^2}$. Thus the full constant is $1/[2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}]$. The same factor that contracts or expands area under the transformation must adjust density in the opposite direction.

Reading the Quadratic Form

The expression $u^2-2\rho uv+v^2$ is the key part of the exponent. Its cross term is $-2\rho uv$, so the sign of $\rho$ matters together with the signs of $u$ and $v$. If $\rho>0$, observations with standardized coordinates of the same sign are less penalized by the quadratic form than opposite-sign coordinates of the same size. If $\rho<0$, the reverse pattern occurs.

The whole quadratic form is divided by $1-\rho^2$. As $|\rho|$ approaches one, this denominator shrinks, reflecting the increasing concentration of the distribution around a line. At $|\rho|=1$, the formula fails: its normalizing constant is undefined, and the two-dimensional model becomes degenerate.

A density value is not itself a probability. Probabilities are obtained by integrating the joint density over a region in the $x,y$ plane. The units of the density are inverse $x$-units times inverse $y$-units, which is another reason to retain both standard deviations in the normalizing constant.

Evaluating the Density Carefully

A reliable calculation separates standardization, the quadratic form, the exponent, and the normalizing constant. This helps prevent the most frequent errors: omitting the cross term, using the wrong sign for it, or forgetting the factor $\sqrt{1-\rho^2}$ in the constant.

Worked Example: Negative Correlation at an Offset Point

Suppose $(X,Y)$ is bivariate normal with $\mu_X=100,\ \sigma_X=10,\ \mu_Y=50,\ \sigma_Y=4,$ and $\rho=-0.5$. Find the joint density at $(x,y)=(110,50)$. Give the answer to four decimal places. Method: standardize first, then evaluate the quadratic form and the density formula.

Standardize. The coordinates are $u=(110-100)/10=1$ and $v=(50-50)/4=0$.

Quadratic form and exponent. The quadratic form is $u^2-2\rho uv+v^2=1^2-2(-0.5)(1)(0)+0^2=1$. Also, $1-\rho^2=1-0.25=0.75$, so the exponent is $-1/[2(0.75)]=-2/3$.

Normalizing constant and density. The constant is $1/[2\pi(10)(4)\sqrt{0.75}]=1/(80\pi\sqrt{0.75})$, approximately $0.004594$. Therefore $f_{X,Y}(110,50)\approx 0.004594e^{-2/3}\approx 0.004594(0.513417)=0.002359$. Rounded to four decimal places, the density is 0.0024.

Check and trap avoided. Since $v=0$, the cross term must be zero. The density is smaller than the normalizing constant because the exponential factor is less than one. Omitting $\sqrt{0.75}$ would give an incorrect constant.

Worked Example: Both Coordinates Away From Their Means

A hypothetical joint model for two measurements has $\mu_X=10,\ \sigma_X=2,\ \mu_Y=12,\ \sigma_Y=4,$ and $\rho=-0.5$. Find the density at $(14,8)$, rounded to four decimal places. Method: calculate both standardized coordinates before applying the correlated quadratic form.

Standardize. Here $u=(14-10)/2=2$, and $v=(8-12)/4=-1$.

Evaluate the quadratic form. The cross term is $-2\rho uv=-2(-0.5)(2)(-1)=-2$. Hence the full form is $u^2-2\rho uv+v^2=4-2+1=3$. The denominator in the exponent is $2(1-\rho^2)=2(0.75)=1.5$, giving exponent $-3/1.5=-2$.

Compute the density. The normalizing constant is $1/[2\pi(2)(4)\sqrt{0.75}]=1/(16\pi\sqrt{0.75})\approx 0.022972$. Thus $f_{X,Y}(14,8)\approx 0.022972e^{-2}\approx 0.022972(0.135335)=0.003109$. Rounded to four decimal places, the density is 0.0031.

Check and trap avoided. The negative correlation and opposite signs of $u$ and $v$ make the cross term negative here. A common error is to treat $-2\rho uv$ as positive just because $\rho$ is negative, without also checking the sign of $uv$.

Worked Example: A Standardized Point With Opposite Signs

Suppose $\mu_X=8,\ \sigma_X=2,\ \mu_Y=20,\ \sigma_Y=3,$ and $\rho=0.5$. Find the density at $(x,y)=(12,17)$, rounded to six decimal places. Method: use the standardized coordinates to isolate the correlation effect.

Standardize. We have $u=(12-8)/2=2$ and $v=(17-20)/3=-1$.

Compute the quadratic form. The cross term is $-2(0.5)(2)(-1)=2$, so $u^2-2\rho uv+v^2=4+2+1=7$. Since $1-\rho^2=0.75$, the exponent is $-7/[2(0.75)]=-7/1.5=-14/3$.

Compute the constant and density. The normalizing constant is $1/[2\pi(2)(3)\sqrt{0.75}]=1/(12\pi\sqrt{0.75})\approx 0.03063$. Consequently, $f_{X,Y}(12,17)\approx 0.03063e^{-14/3}\approx 0.03063(0.0094036)=0.000288$. To six decimal places, the density is 0.000288.

Check and trap avoided. The standardized coordinates have opposite signs while the correlation is positive, so the cross term increases the quadratic form from $5$ to $7$, reducing the exponential factor. Replacing the quadratic form with $u^2+v^2$ would miss this effect.

Worked Example: A High Positive Correlation

A hypothetical pair has $\mu_X=3,\ \sigma_X=1.5,\ \mu_Y=5,\ \sigma_Y=2,$ and $\rho=0.8$. Find the density at $(4.5,5)$, rounded to five decimal places. Method: retain the correlation term even though one coordinate is at its mean.

Standardize. The values are $u=(4.5-3)/1.5=1$ and $v=(5-5)/2=0$. Thus the quadratic form is $1^2-2(0.8)(1)(0)+0^2=1$.

Evaluate the exponent. Since $1-\rho^2=1-0.64=0.36$, the exponent is $-1/[2(0.36)]=-1/0.72=-1.3888889$, using the unrounded value in the next calculation.

Evaluate the constant. It is $1/[2\pi(1.5)(2)\sqrt{0.36}]=1/[2\pi(3)(0.6)]=1/(3.6\pi)\approx 0.0884194$. Therefore $f_{X,Y}(4.5,5)\approx 0.0884194e^{-1.3888889}\approx 0.0884194(0.2493522)=0.0220476$. Rounded to five decimal places, the density is 0.02205.

Check and trap avoided. With $v=0$, the cross term vanishes, but $\rho$ still appears in $1-\rho^2$ in both the exponent and the constant. Setting the cross term to zero does not mean the density can be computed as if the variables were independent.

Common Exam Traps

The derivation suggests a short checklist for calculations. First confirm that the model is bivariate normal and nondegenerate. Then standardize using each variable’s own mean and standard deviation. Next evaluate the complete quadratic form, keeping its cross term intact, and only then calculate the exponent and constant. Keep full precision through the multiplication; round the final density only as requested.

Using raw deviations in the quadratic form. The symbols $u$ and $v$ are standardized coordinates, not $x-\mu_X$ and $y-\mu_Y$.
Dropping or mis-signing the cross term. Its value is $-2\rho uv$; the signs of all three quantities matter.
Using an independent-normal constant. The nonzero-correlation constant includes $\sqrt{1-\rho^2}$, in addition to $2\pi\sigma_X\sigma_Y$.
Calling a density a point probability. For a continuous pair, the probability at a single point is zero; a density is a local rate of probability per unit area.
Substituting $|\rho|=1$. The formula is for $-1<\rho<1$; at the endpoints the joint distribution is concentrated on a line.

The quadratic form and the Jacobian explain the full structure of the density: correlation changes the exponent through the cross term and changes the area scaling through the square root factor. These are the core ingredients needed to derive or evaluate the bivariate normal density, aligned with the SOA Exam P (Probability) syllabus.

Check Your Understanding

For each problem, standardize first and keep enough precision to round only at the final step.

A nondegenerate pair has standard deviations 2 and 3 and correlation zero. At the point where both variables equal their respective means, find the density to four decimal places. (Answer: 0.0265)
A pair has $\mu_X=10,\sigma_X=2,\mu_Y=5,\sigma_Y=1,\rho=0.5$. Find the density at $(12,6)$ to four decimal places. (Answer: 0.0472)
A pair has $\mu_X=0,\sigma_X=3,\mu_Y=10,\sigma_Y=2,\rho=-0.4$. Find the density at $(3,6)$ to four decimal places. (Answer: 0.0038)
A pair has $\mu_X=4,\sigma_X=2,\mu_Y=7,\sigma_Y=5,\rho=0.6$. Find the density at $(5,4.5)$ to five decimal places. (Answer: 0.01065)
A model has positive standard deviations and correlation $\rho=1$. Explain why the nondegenerate density formula cannot be used, and state the geometric form of the support. (Answer: The formula's factor $\sqrt{1-\rho^2}$ is zero, so there is no ordinary two-dimensional density; the distribution is concentrated on the line $Y=\mu_Y+(\sigma_Y/\sigma_X)(X-\mu_X)$.)

What You'll Learn